You have found the following ages (in years) of 5 gorillas. Those gorillas were randomly selected from the 31 gorillas at your local zoo: $ 24,\enspace 9,\enspace 27,\enspace 14,\enspace 21$ Based on your sample, what is the average age of the gorillas? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 31 gorillas, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{24 + 9 + 27 + 14 + 21}{{5}} = {19\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {25} + {100} + {64} + {25} + {4}} {{5 - 1}} $ {s^2} = \dfrac{{218}}{{4}} = {54.5\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{54.5\text{ years}^2}} = {7.4\text{ years}} $ We can estimate that the average gorilla at the zoo is 19 years old. There is also a standard deviation of 7.4 years.